Representations of the (-2,3,7)-Pretzel Knot and Orderability of Dehn Surgeries

Konstantinos Varvarezos November 27, 2019

Abstract

We construct a 1-parameter family of SL2(R) representations of the pretzel knot P (−2, 3, 7). As a consequence, we conclude that Dehn surgeries on this knot are left-orderable for all rational surgery slopes less than 6. Furthermore, we discuss a family of knots and exhibit similar orderability results for a few other examples.

1 Introduction

This paper studies the character variety of a certain knot group in view of

the relationship between PSL^2(R) representations and the orderability of Dehn surgeries on the knot. This is of interest because of an outstanding conjectured relationship between orderability and L-spaces. A left-ordering on a group G is a total ordering ≺ on the elements of G that is invariant under left-multiplication; that is, g ≺ h implies fg ≺ fh for all f, g, h ∈ G. A group is said to be left-orderable if it is nontrivial and admits a left ordering. A 3-manifold M is called orderable if π1(M) is arXiv:1911.11745v1 [math.GT] 26 Nov 2019 left-orderable. If M is a rational homology 3-sphere, then the rank of its Heegaard Floer homology is bounded below by the order of its ﬁrst (integral) homology group. M is called an L-space if equality holds; that is, if rk HFd(M) = |H1(M; Z)|. This work is motivated by the following proposed connection between L-spaces and orderability, ﬁrst conjectured by Boyer, Gordon, and Watson.

1 Conjecture 1.1 ([BGW13]). An irreducible rational homology 3-sphere is an L-space if and only if its fundamental group is not left-orderable. In [BGW13], this equivalence was shown to hold for all closed, connected, orientable, geometric three-manifolds that are non-hyperbolic. Knot groups are of particular interest due to the fact that knot complements are often hyperbolic and also because the L-space surgery interval for an L-space knot is known to be [2g − 1, ∞), where g is the Seifert genus of the knot [OSz11]. We shall primarily focus on the (−2, 3, 7)-pretzel knot, which is an L- space knot. This knot has genus 5, and so if Conjecture 1.1 holds, one would expect the orderable surgeries to be precisely from slopes in the interval (−∞, 9). It has been shown in [Nie19] that surgery along any slope outside this interval always yields a non-orderable manifold. Moreover, Nie also showed that surgery along a slopes in (−, ) yields an orderable manifold for some sufficiently small > 0. In this work, we improve the result to: Theorem 1.2. Let X denote the exterior of the (−2, 3, 7)-pretzel knot. Then X(r) is orderable for all r ∈ (−∞, 6). This leaves the slope interval [6, 9) as still unverified vis-à-visConjecture 1.1. The proof of the main theorem follows the strategy developed by Culler and Dunfield in [CD18]. In particular, we compute a certain one-parameter family of PSL^2(R)-representations of the knot group, and from that, we conclude the existence of a curve on the translation extension locus associated to such representations. In fact, Culler and Dunfield used numerical methods to produce an image of the translation extension locus associated to the knot P (−2, 3, 7) (see Figure 3 in [CD18]). Theorem 1.2 is precisely the result one expects from that image. The (−2, 3, 7)-pretzel knot can be viewed as one of a family of twisted torus knots. In particular, consider the (3, 3k + 2) torus knot with m full twists about a pair of strands (pictured in Figure 1), which we denote by m 1 T3,3k+2. Notice that T3,5 is the pretzel knot P (−2, 3, 7). In section 5, we 1 consider possible extensions of our result to the family T3,3+2.

Acknowledgements The author would like to thank Professors ZoltánSzabóand Peter Ozsváth for suggesting and encouraging work on this problem. This work was sup- ported by the NSF RTG grant DMS-1502424.

2 k m

m Figure 1: The twisted torus knot T3,3k+2. Here the boxes represent k and m full twists of the strands passing through.

2 Background ˜ In what follows, we shall denote by G the group PSL2(R) and by G we shall denote its universal cover PSL^2(R).

2.1 Reducible Representations Let K ⊆ S3 be a knot and denote by X the knot exterior. The abelianization ∼ map π1(X) → H1(X; Z) = Z sends the meridian µ to a generator of the ﬁrst homology. For any ζ ∈ S1 ⊆ C (the complex unit circle), one can deﬁne a ∼ reducible representation ρζ : π1(X) → PSU(1, 1) = G given by composing the abelianization map with:

H1(X; Z) → PSU(1, 1) ζ1/2 0 [µ] 7→ ± 0 ζ−1/2 where ζ1/2 is any square root of ζ.

3 We are interested in ﬁnding paths of irreducible representations that are deformations of some such reducible representation. The following theorem gives necessary and suﬃcient conditions for this to happen:

Theorem 2.1 (Lemma 4.8 and Proposition 10.3 of [HP05]). With K and ζ as above, ρζ may be deformed into a family of irreducible PSL2(C) representations if and only if ζ is a root of the Alexander polynomial. In that case, the irreducible curve of characters is contained in PSU(2) on one side and ∼ in PSU(1, 1) = PSL2(R) on the other side.

2.2 The Translation Extension Locus Following [CD18], we give a description of the translation extension locus and ˜ explain how it may be used to construct left-orderings. Let ρ : π1(X) → G be a representation of the fundamental group of the knot exterior. As G acts on the circle via orientation-preserving transformations, G˜ acts on R via orientation-preserving transformations as well and, in fact, can be viewed as a subgroup of Homeo+(R). More precisely, we are viewing R as the universal cover of S1 via the covering map x 7→ e2πix so that, for instance, lifts of the identity matrix act on R by integral translations. We are interested in when such representations factor through the fundamental group of the ﬁlled manifold X(r), for then, by [BRW05], it will follow that X(r) will be left-orderable. Let us deﬁne the translation map trans : G˜ → R by

gn(x) − x trans(g) := lim n→∞ n for any x ∈ R (note this deﬁnition does not depend on x). We call an element of G˜ elliptic, parabolic, or hyperbolic if its image under the natural projection p : G˜ → G is elliptic, parabolic, or hyperbolic, respectively. Note ˜ 1 that for any elliptic g ∈ G, trans(g) is equal to 2π times the rotation angle corresponding to the action of p(g), up to additon of an integer. Now let us deﬁne the translation extension locus. For each representation ˜ ρ : π1(X) → G, we may consider its restriction to the peripheral subgroup: ˜ ∼ 2 ρ|π1(∂X) : π1(∂X) → G. As ∂X is a torus, π1(∂X) = Z is abelian, and hence all (nontrivial) elements in the image of ρ|π1(∂X) are of the same type (elliptic, parabolic, or hyperbolic). So we shall call ρ elliptic, parabolic, or hyperbolic if the image of the restriction to the peripheral subgroup consists

4 of elements of the respective type. Given ρ as above, we may consider the composition trans ◦ρ|π1(∂X) : π1(∂X) → R. Because the peripheral subgroup is abelian, this is actually a homomorphism, and hence may be considered ∼ 1 ∼ 2 as an element of Hom(π1(∂X), R) = H (∂X, R) = R . If (µ, λ) are the natural meridian-longitude basis for H1(∂X, R), then one may consider the dual basis (µ∗, λ∗) for H1(∂X, R). Using this basis, we may consider the subset E of R2 corresponding to all elliptic and parabolic representations ˜ ρ : π1(X) → G. The translation extension locus is the closure of this set. We ¯ 2 denote it ELG˜(X) := E ⊆ R . p Let r = q ∈ Q with p, q relatively prime. Then γ = pµ + qλ is a primitive element of H1(∂X, Z) and is represented by a simple closed curve 1 in ∂X. Let Lr denote the set of elements of H (∂X, R) which vanish on γ. In the meridian-longitude-dual basis, this corresponds to the (a, b) ∈ R2 such that ap + bq = 0. Graphically, this is the line through the origin in 2 p R with slope − q = −r. Suppose an element in ELG˜(X) ∩ Lr comes from an elliptic representation ρ. Then, trans(ρ(µpλq)) = 0. As remarked above, the translation of an elliptic element corresponds to the rotation angle about its fixed point, and so, ρ(µpλq) = 1 ∈ G.˜ It follows that ρ factors through p q ∼ π1(X)/hhµ λ ii = π1(X(r)). It follows by [BRW05] that X(r) is orderable if it is irreducible. In fact, Culler and Dunfield proved the following slightly more general result, for M any rational homology solid torus: Theorem 2.2 (Lemma 4.4 of [CD18]). Suppose M is a compact orientable irreducible 3-manifold with ∂M a torus, and assume the Dehn filling M(r) is irreducible. If Lr meets ELG˜(M) at a nonzero point which is not parabolic or ideal, then M(r) is orderable. Remark. Here an ideal point is an element of E¯ \ E so that the hypotheses of the theorem may be equivalently expressed as Lr intersecting ELG˜(M) at a nonzero elliptic point. Culler and Dunfield show that the translation extension locus has the following properties:

Theorem 2.3 (Theorem 4.3 of [CD18]). The extension locus ELG˜(M) is a locally finite union of analytic arcs and isolated points. It is invariant under D∞(M) with quotient homeomorphic to a finite graph. The quotient contains finitely many points which are ideal or parabolic in the sense defined above. The locus ELG˜(M) contains the horizontal axis Lλ, which comes from representations to G˜ with abelian image.

5 2 Here D∞(M) is a copy of the inﬁnite dihedral group which acts on R via horizontal translations and rotations which depend on the homology of M. In the case that M is a knot exterior, D∞(M) is generated by the unit horizontal translation (x, y) 7→ (x + 1, y) and the rotation about the origin (x, y) 7→ (−x, −y). In particular, the section of the translation extension 1 locus with 0 ≤ x ≤ 2 is a fundamental domain for the entire locus.

2.3 The Knot Group As noted in the inroduction, the knot K can be interpreted as a twisted torus knot. We shall use a presentation of the group π1(X) that comes from this interpretation, citing a result of Clay and Watson.

m Lemma 2.4 ([CW10]). If X is the exterior of the twisted torus knot T3,3k+2, then 2 −k m 2k+1 −k m k+1 π1(X) = a, b a b a a = b b a b Moreover, the peripheral subgroup is generated by the meridian µ and longitude σ given in this presentation by:

µ = a−1bk+1 m m σ = a b−ka a b−ka a

The homological longitude λ is then

λ = µ−3(3k+2)−4mσ

Remark. This is essentially the content of Propositions 3.1 and 3.2 in [CW10] except that there, conjugates of µ and σ by the generator a are used instead. Also note that there is a misprint in the paragraph following Proposition 3.2 in the formula for the homological longitude. Indeed, by considering the abelianization, one ﬁnds that: [a]3 = [b]3k+2 and moreover, [a] = [µ]3k+2 and [b] = [µ]3. As noted in the introduction, when k = 1 we recover the pretzel knots P (−2, 5, 2m + 5). So setting m = 1 as well, we obtain:

Corollary 2.5. If X is the exterior of the pretzel knot P (−2, 3, 7), then

2 −1 2 2 2 π1(X) = a, b a b a = b ab

6 Moreover, the peripheral subgroup is generated by the meridian µ and longitude σ given in this presentation by:

µ = a−1b2 σ = ab−1a2b−1a2

The homological longitude λ is then

λ = µ−19σ

3 Computing Representations

In our computation of representations, we follow the methods of [Che18], who computed the SL2(C) character variety for even 3-stranded pretzel knots. For our purposes, it is more convenient to apply those methods to this particular presentation rather than using his result directly. We state here the relevant facts:

Lemma 3.1. If X,Y ∈ SL2 C then (a) XYX = tr(XY )X − Y −1;

(b) tr(X−1) = tr(X);

k (d) tr(X ) = ωk(tr(X))X − ωk−1(tr(X))I; (e) X and Y have no common eigenvector if and only if I,X,Y,XY are linearly independent.

Here I is the 2 × 2 identity matrix, and the ωk are polynomials charac- terized by:

ω0(t) ≡ 0

ω1(t) ≡ 1

ωk+1(t) = tωk(t) − ωk−1(t)

ω−k(t) = −ωk(t)

7 In particular, we have:

X2 = tr(X)X − I X−1 = tr(X)I − X

Let ρ : π1(X) → SU(1, 1) be a representation for the knot group. Put A = ρ(a) and B = ρ(b), where a and b are the generators of the group as in the presentation in Lemma 2.5. Let us also put t = tr(A), s = tr(B), and r = tr(AB). Using the lemma, we may compute:

A2B−1A2 = tr(A2B−1)A2 − (B−1)−1 = tr((tA − I)(sI − B))(tA − I) − B = (ts · tr(A) − s · tr(I) − t · tr(AB) + tr(B))(tA − I) − B = (t2s − tr − s)(tA − I) − B = (t2s − tr − s)tA − (t2s − tr − s)I − B (1)

Similarly,

B2AB2 = tr(B2A)B2 − A−1 = tr((sB − I)A)(sB − I) − (tI − A) = (s · tr(BA) − tr(A))(sB − I) − tI + A = (rs − t)(sB − I) − tI + A = (rs − t)sB − rsI + A (2)

3.1 Irreducible Representations According to the group presentation in Corollary 2.5, we must have that A2B−1A2 = B2AB2, and hence, by Lemma 3.1 (e), if ρ is an irreducible representation then the equality of (1) and (2) implies that:  1 = (t2s − tr − s)t  −1 = (rs − t)s (3) rs = (t2s − tr − s) This can be solved in terms of t, and hence we have the following:

8 Lemma 3.2. If ρ : π1(X) → SU(1, 1) is an irreducible representation with tr(ρ(a)) = t, tr(ρ(b)) = s, and tr(ρ(ab)) = r then

( t s = t2−1 1 (4) r = 1 − t2 Now let us examine the “intersection” between reducible and irreducible representations; this will correspond to the points at which reducible representations may be deformed to irreducible ones. In the notation of section 2.1, the traces of reducible representations satisfy:

tr(A) = e5iθ + e−5iθ tr(B) = e3iθ + e−3iθ tr(AB) = e8iθ + e−8iθ

Substituting z = eiθ and t, s, r for the traces gives:

t = z5 + z−5 s = z3 + z−3 r = z8 + z−8

Hence, by (4): ( 3 −3 z5+z−5 z + z = (z5+z−5)2−1 8 −8 1 z + z = 1 − (z5+z−5)2 which is equivalent to: ( z13 + z−7 + z3 + z7 + z−13 + z−3 − z5 − z−5 = 0 z18 + z−2 + 2z8 + z2 + z−18 + 2z−8 − z10 − z−10 − 1 = 0 which one can factor as: ( (z−1 + z)(z−2 + z2)(z−10 − z−8 + z−4 − z−2 + 1 − z2 + z4 − z8 + z10) = 0 (z−8 + z−6 + z−4 − 1 + z4 + z6 + z8)(z−10 − z−8 + z−4 − z−2 + 1 − z2 + z4 − z8 + z10) = 0

One sees that z2 must be a root of the polynomial x−5 −x−4 +x−2 −x−1 +1− x + x2 − x4 + x5, which is the Alexander polynomial of the knot P (−2, 3, 7). This is consistent with the general result of Theorem 2.1, which implies that

9 reducible representations can be deformed into irreducible ones only if z2 is a root of the Alexander polynomial. Let us consider consider a deformation of the representation correspond- iθ0 ing to the root x = e with θ0 ≈ 2.0453 ≈ 0.3255(2π). This corresponds to 5θ0 t0 = 2 cos( 2 ) ≈ 0.78

3.2 Realization of Representations Let us investigate a partial converse of the previous lemma. In particular, given real numbers t, s ,and r, when does there exist a representation ρ : π1(X) → SU(1, 1) satisfying tr(ρ(a)) = t, tr(ρ(b)) = s, and tr(ρ(ab)) = r? Let us restrict ourselves to the case where ρ(a) is elliptic. Then, we may suppose, by conjugating appropriately if necessary, that: eiα 0 ρ(a) = 0 e−iα √ 1 + R2eiβ R ρ(b) = √ R 1 + R2e−iβ √ 1 + R2ei(α+β) Reiα ρ(ab) = √ Re−iα 1 + R2e−i(α+β) for some α, β, R ∈ R. Hence we are looking for real solutions of: t = tr(ρ(a)) = 2 cos α √ s = tr(ρ(b)) = 2 1 + R2 cos β √ r = tr(ρ(ab)) = 2 1 + R2 cos(α + β) For s 6= 0 we can divide the last two equations: r cos(α + β) = s cos β = cos α − sin α tan β As |t| < 2, cos α 6= 1 so that sin α 6= 0. Hence, under these conditions: t cos α = 2 1 t r tan β = − sin α 2 s

10 which determine α and β (up to sign or multiples of π). Finally, we can see that such a representation can be constructed as long as there is an R satisfying: s 2 R2 = − 1 (5) 2 cos β which happens exactly when

s ≥ 1. (6) 2 cos β Notice that when the above inequality fails, one can ﬁnd a purely imag- inary R that satisﬁes the condition (5). In this case, the representation is actually in SU(2) instead. This is consistent with Theorem 2.1, which implies that deformations of reducible representations will be in SU(1, 1) on one “side” and in SU(2) on the other. If we further suppose t, s, r satisfy condition 4, then we see that 0 < t < 1 implies s 6= 0 and that s varies con- tinuously. Therefore, for a continuous path of t-values within this interval, one can produce a continuous path of representations into SU(2) ∪ SU(1, 1).

4 Constructing a Path in the Translation Ex- tension Locus

4.1 Trace Computations Here we use the calculations of the previous section to demonstrate the existence of a certain path in the translation extension locus. Notice that, although the previous section considered representations into SU(1, 1), that group is conjugate to SL2(R). Moreover, it suﬃces to produce a path of representations into SL2(R) as such a path will lift to a path of representations into G,˜ the universal cover. By Corollary 2.5, ρ(µ) = A−1B2. Hence, we compute:

tr(ρ(µ)) = tr((tI − A)(sB − I)) = ts · tr(B) − t · tr(I) − s · tr(AB) + tr(A) = ts2 − sr − t

11 Similarly, we have that ρ(σ) = AB−1A2B−1A2, and hence, using (1), tr(ρ(σ)) = tr((AB−1)((t2s − tr − s)tA − (t2s − tr − s)I − B)) = (t2s − tr − s)t · tr(AB−1A) − (t2s − tr − s) tr(AB−1) − tr(A) = (t2s − tr − s)t · tr(tr(AB−1)A − (B−1)−1) − (t2s − tr − s) tr(AB−1) − tr(A) = (t2s − tr − s)t · tr(tr(A(sI − B))A − B) − (t2s − tr − s) tr(A(sI − B)) − tr(A) = (t2s − tr − s)t · tr((st − r)A − B) − (t2s − tr − s)(st − r) − t = (t2s − tr − s)t((st − r)t − s) − (t2s − tr − s)(st − r) − t = (t2s − tr − s)2t − (t2s − tr − s)(st − r) − t

Using the conditions in (3) and (4), these may be expressed in terms of t:

t3 1 tr(ρ(µ)) = − − t =: m(t) (7) (t2 − 1)2 t 2 t 1 tr(ρ(σ)) = − − − t =: `(t) (8) t t2 − 1 t3 Lemma 4.1. For t ∈ (0, 1), m(t) and `(t) are strictly increasing as functions of t. Proof. One computes, by (7):

3t2(t2 − 1) − 4t4 1 m0(t) = + − 1 (t2 − 1)3 t2 The ﬁrst term on the right-hand-side is seen to be positive for positive t < 1, 1 and t2 > 1 for such values of t as well. Similarly, by (8),

2 t2 − 1 − 2t2 3 `0(t) = − − + − 1 t2 (t2 − 1)2 t4 1 + t2 (t2 + 3)(1 − t2) = + (t2 − 1)2 t4 which is seen to be positive for 0 < t < 1. Lemma 4.2. If s and r satisfy condition (4), then on the interval t ∈ √ √ p p t r 2 − 2, 10 − 2 the quantity 2 − s is positive and increasing as a function of t.

12 Proof. It is straightforward to compute that

t r t 2 1 − = − − + 2 s 2 t t3 (t2 − 2)2 − 2 = − 2t3 √ √ which is easily seen to be positive whenever − 2 < t2 − 2 < 2, and so in p √ p √ particular, it is positive when t ∈ 2 − 2, 2 + 2 . Now, diﬀerentiat- ing with respect to t gives

d t r 1 2 3 − = − + − dt 2 s 2 t2 t4 (t2 + 2)2 − 10 = − 2t4 √ √ which is positive whenever − 10 < t2 + 2 < 10, and this is satisﬁed p√ when t ∈ 0, 10 − 2 . The interval in the statement of the lemma is the intersection of these two intervals.

4.2 A Path of Irreducible Representations

Now observe that, on the interval t ∈ (t0, t1), α is increasing as a function 3π of t since α remains between 2 and 2π. This implies that sin α is negative and increasing on this interval. Hence, by lemma 4.2, tan β is negative and decreasing on this interval. Because β is in the second quadrant, it follows t that cos β is negative and increasing. Notice that s = t2−1 is negative and s decreasing on this interval (since t1 < 1), and so the quantity 2 cos β is positive and increasing on the interval. Therefore, condition (6) is satisﬁed on the interval, and we have shown:

Theorem 4.3. There exists a continuous path of elliptic representations ρ(t): π1(X) → SU(1, 1) for t ∈ [t0, t1) which is irreducible except at t0 and is such that ρ(t0) = ρ0 and tr(ρ(t)(a)) = t. Now observe that, since µ and σ commute, if ρ is elliptic then ρ(µ) and ρ(σ) have the same ﬁxed point. Hence, up to conjugation, they have the

13 form: eiφ 0 ρ(µ) = 0 e−iφ eiψ 0 ρ(σ) = 0 e−iψ

φ So that tr(ρ(µ)) = 2 cos φ and tr(ρ(σ)) = 2 cos ψ , and also trans(ρ(µ)) = π . Moreover,

ρ(λ) = ρ(µ−19σ) ei(−19φ+ψ) 0 = 0 ei(19φ−ψ)

−19φ+ψ Hence trans(ρ(λ)) = π . So the point on the translation extension locus φ −19φ+ψ ∼ is: ( π , π ). Note that, when considering the above matrices in SU(1, 1) = SL2(R), the values of φ and ψ are only relevant modulo 2π. However, since the extension translation locus is constructed from lifted representations to G,e it will be important to keep track of the actual values. Proof of Theorem 1.2. Let us investigate the part of the extension translation locus coming from the path of representations given by Theorem 4.3. θ0 At t = t0, we begin with a reducible representation with coordinates ( 2π , 0). Hence, at this point, φ ≈ 1.02 and ψ ≈ 19.38 ≈ 6π + .53. Using the notation 5θ0 3θ0 of section 3.2, we have that, at this point, α = 2 ≈ 5.11 and β = 2 ≈ 3.07. As t → t1, lemma 4.1 tells us that m(t) increases monotonically to 2, and hence φ decreases monotonically to 0. Similarly, `(t) also increases monotonically to 2 so that ψ decreases monotonically to 6π. Therefore, the limiting parabolic point in the translation extension locus is (6, 0), and in fact there is θ0 a path in the translation extension locus connecting ( 2π , 0) and (6, 0) which ∗ ∗ 2 ∗ θ0 is contained within (µ , λ ) ∈ R : 0 ≤ µ ≤ 2π (see Figure 2). By the symmetries of the translation extension locus (Theorem 2.3), there is also a θ0 continuous path connecting the points (1− 2π , 0) and (1, −6) contained within ∗ ∗ 2 θ0 ∗ (µ , λ ) ∈ R : 1 − 2π ≤ µ ≤ 1 . It is easy to see that all lines through the origin of slope greater than -6 will intersect one of these paths. By construc- tion, these paths consist of elliptic points, and so, by Theorem 2.2, the Dehn ﬁlling M(r) for r ∈ (−∞, 6) will be orderable whenever it is irreducible. As the exceptional slopes of the knot K, as given in Problem 1.77 of [Kir96] all

14 6

0.05 0.10 0.15 0.20 0.25 0.30

Figure 2: The translation extension locus for the 1-parameter family of representations constructed above. lie outside this interval, all Dehn ﬁllings within the interval are hyberbolic and hence irreducible. Therefore, they are all orderable.

Remark. Figure 2 shows a plot of the part of the translation extension locus corresponding to the path in Theorem 4.3, computed using Mathematica [Wol]. One sees that the curve in the extension translation locus actually remains within the ﬁrst quadrant and resembles a straight line (as did the analogous curve in [CD18]. However, explicit computation shows that it is not actually a line. Indeed, ﬁgure 3 shows how the slope of this curve varies with the parameter t. It is interesting to note that it ranges between -18.4 −19φ+ψ to -18.5, suggesting that the meridional (−19φ) term in π is dominant, with the “deviation” coming from ψ contributing less than 1 to the slope. Indeed, the joint monotonicity of m(t) and `(t), as given by Lemma 4.1 essentially shows that the total “deviation” of the endpoint from that of a line of slope -19 must be less than one, as the total displacement of ψ must be less than one. Also using Mathematica, one can plot the translation extension locus of this knot coming from all roots of the Alexander polynomial (hence, giving a picture of two fundamental domains of the entire locus). This is shown in Figure 4, and it agrees with the corresponding diagram obtained by Culler

15 18.44

18.43

18.42

18.41

0.785 0.790 0.795 0.800 Figure 3: The slope of the curve of the translation extension locus constructed in the proof of Theorem 1.2, as a function of t. and Dunfield’s (see Figure 3 of [CD18]; the reason for the vertical reflection is most likely a difference in convention for the orientation of the meridian- longitude pair).

5 Generalizations

5.1 A Family of Twisted Torus Knots

1 We now consider the family of twisted torus knots T3,3k+2 for k ≥ 1, which contains the (−2, 3, 7)-pretzel knot as the special case k = 1. By Lemma 2.4, the fundamental group of the knot exterior has presentation:

1 2 −k 2 k+1 k+1 π3,3k+2 := a, b a b a = b ab

1 As in section 3, we see that if a representation ρ : π3,3k+2 → SL2(R) is such that ρ(a) = A and ρ(b) = B, then using Lemma 3.1, we see that:

A2B−kA2 = tr(A2B−k)A2 − Bk 2 −k = tr(A B )(tA − I) − (ωk(s)B − ωk−1(s)I) (9)

16 6

0 0.2 0.4 0.6 0.8 1.0

-2

-4

-6

Figure 4: A (double) fundamental domain of the translation extension locus for the (−2, 3, 7)-pretzel knot where t = tr(A), s = tr(B), and r = tr(AB). Similarly:

Bk+1ABk+1 = tr(Bk+1A)Bk+1 − A−1 k+1 = tr(ωk+1(s)B − ωk(s)I)B − (tI − A)

= (rωk+1(s) − tωk(s))(ωk+1(s)B − ωk(s)I) − tI + A (10)

Equating (9) and (10), we see by Lemma 3.1(e), if ρ is irreducible then:  tr(A2B−k) = 1  t −ωk(s) = (rωk+1(s) − tωk(s)) (11)  2 −k − tr(A B ) + ωk−1(s) = −(rωk+1(s) − tωk(s))ωk(s) − t

From this we readily ﬁnd that the following must hold:

2 ( 1 ωk(s) t − = −ωk−1(s) + t ωk+1(s) (12) tωk(s) ωk(s) r = − 2 ωk+1(s) ωk+1(s) Since the irreducible characters are one-dimensional, it follows that these must be all the relations. Notice also that, as long as ωk+1(s) 6= 0, each value of s corresponds to exactly two values of t (each being the negative reciprocal of the other) and hence two values of r.

17 We may write the traces of the images of the meridian and (surface- framed) longitude in terms of these parameters. By Lemma 2.4, µ = a−1bk+1 and so:

tr(ρ(µ)) = tr(A−1Bk+1) = tr((tI − A)Bk+1)

= t(sωk+1(s) − 2ωk(s)) − (rωk+1(s) − tωk(s))

ωk(s) = t(sωk+1(s) − 2ωk(s)) + ωk+1(s) where the condition (12) was used in the last step. Similarly, by Lemma 2.4, we know that σ = ab−ka2b−ka2 so that: tr(ρ(σ)) = tr(AB−kA2B−kA2) = tr(AB−k(tr(A2B−k)A2 − Bk)) = tr(A2B−k) tr(AB−kA2) − tr(A) = tr(A2B−k) tr((tr(AB−k)A − Bk)A) − t = tr(A2B−k)(tr(AB−k) tr(A2) − tr(BkA)) − t 2 −k 2 = tr(A B )(ω−k(s)r − ω−k−1(s)t)(t − 2) − (ωk(s)r − ωk−1(s)t)) − t 1 = (ω (s)t − ω (s)r)(t2 − 2) − (ω (s)r − ω (s)t)) − t t k+1 k k k−1 where the condition (11) was used in the last step.

5.2 Empirical Observations Using the calculations of the previous section, one can compute the transla- 1 tion extension locus corresponding to the twisted torus knot T3,3k+2 for any value of k. Indeed, by [Mor06], we have an explicit formula for the Alexander 1 polynomial of T3,3k+2, from which the roots may be approximated. Moreover, for each deformation coming from a root of the Alexander polynomial, we can verify that equation (6) holds (or the analogue with the roles of s and t reversed), hence conﬁrming that these come from SL2(R) representations. Figure 5 shows the graphs of these extension translation loci for k = 2, 3, 4, produced using Mathematica. From these we can make the following observations:

18 10 15

10 5 5 5

0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0

-5 -5 -5 -10

-10

-10 -15 (a) k = 2 (b) k = 3 (c) k = 4

Figure 5: Double fundamental domains of extension translation loci for the 1 knots T3,3k+2 for k = 2, 3, 4.

1 (i) The translation extension locus corresponding to T3,3k+2 seems to consist of several nearly parallel “almost lines”, each with a “slope” approximately 3(3k + 2) + 4.

(ii) The “lines” corresponding to roots of the Alexander polynomial with argument in (0, π) seem to lie above the horizontal axis (hence, by the symmetries of the translation extension locus, those with argument in (π, 2π) lie below the axis).

(iii) The longest “line” seems to correspond to a root of the Alexander 2π polynomial with argument approximately 3 . (iv) The maximum height of the translation extension locus seems to be achieved by the point (0, 3k + 3)

(i) was also noticed by Culler and Dunfield in [CD18], who asked if translation extension loci for all twisted torus knots (and also Berge knots) have this form. Notice that (iv) would imply that all surgery slopes in the interval (−∞, 3k+ 3) yield orderable manifolds (as long as they are irreducible). On the other 1 hand, notice that the Seifert genus of T3,3k+2 is 3k + 2. By the result of [OSz11], the non-L-space surgeries are those in the interval (−∞, 6k + 3). Moreover, Tran has shown that surgeries in that interval do in fact yield non-orderable manifolds [Tra19]. Hence, it seems that this method of constructing left-orderings leaves out the slopes in the interval [3k + 3, 6k + 3) as still unconfirmed vis-à-visConjecture 1.1. Moreover, (iv) is roughly a consequence of (i) and (iii) since if those two 1 hold, then one would expect the longest “line” which starts near ( 3 , 0) to

19 4 reach the vertical axis near (0, 3k + 2 + 3 ). The nearest point on the integer lattice is (0, 3k+3) (notice that a representation corresponding to such a point is parabolic, and parabolic elements of G˜ must have integer translation). More precisely, suppose the analogue of Lemma 4.1 holds for an appro- priate interval (in particular, an interval corresponding to one of the “lines”). Using the fact that points on the translation extension locus have the form φ −(3(3k+2)+4)φ+ψ ( π , π ), one sees that if the “line” has one endpoint at (x, 0), then ψ = (3(3k+2)+4)πx at that endpoint. At the other endpoint, φ = 0, so that ψ the endpoint will be (0, π ), where ψ ≥ ((3(3k +2)+4)x−1)π, by monotonicity. Now if (iii) holds, the longest “line” has one endpoint at ≈ (1/3, 0) so that the maximal height is at least 3k + 2 + 1/3 (and similarly, monotonicity implies this height will be at most 3k + 4 + 1/3. As mentioned above, the height must be an integer, and so obtain the maximal height either 3k + 3 or 3k + 4. For k ≥ 2, it is not the case that the meridional and longitudinal traces are monotonic on the same intervals. Nevertheless, for the examples computed above, an analogue of Lemma 4.1 does appear to hold for all the intervals corresponding to “lines” in the translation extension locus.

References

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